Hi, I am currently investigating the Collatz conjecture out of curiosity, having watched a video made about it by Numberphile.

My current understanding of the problem is that any odd $n$, written as $2^{a_s}(2b_s-1)-1$, will iterate to even $3^{a_s}(2a_s-1)-1$ after $a_s$ iterations of $\frac{3n+1}{2}$, then being divided by 2 until it becomes some odd $2^{a_{s+1}}(2b_{s+1}-1)-1$.

When $3^a\left(2b-1\right)-1$ is written as $3^ab-\left(3^a+1\right)$, factoring out the highest power of 2 shows that for $a \equiv b \pmod 2$, the highest power of 2 dividing $3^a\left(2b-1\right)-1$ is 2, and for $a \equiv 1 \pmod 2$ where $b \equiv 0 \pmod 4$, or $a \equiv 0 \pmod 2$ where $b \equiv 3 \pmod 4$, it must be 4. In other cases it seems to be no longer dependant on $a$ or $b$ and becomes pseudo-random.

If the Collatz conjecture is true, for any combination of $(a_1,b_1)$ there must be some integer $z$ such that both $a_z$ and $b_z$ equal 1 after $z$ of iterations of $2^{a_s}(2b_s-1)-1 \rightarrow 2^{a_{s+1}}(2b_{s+1}-1)-1$.

An expansion of the $z$th iteration of $2^{a_s}(2b_s-1)-1 \rightarrow 2^{a_{s+1}}(2b_{s+1}-1)-1$ on some odd $n=2^{a_1}(2b_1-1)-1$, $T_z(n_1)$, implies that:

$$T_z(n_1)=\frac{1.5^{\sum_{c=2}^{z}a_c}\left(1.5^ {a_1}\left(n_1+1\right)-1\right)}{2^{\sum_{d=1}^{z}b_d}}+\frac{1.5^{a_z}-1}{2^{b_z}}+\sum_{e=2}^{z-1}\frac{1.5^{\sum_{f=e+1}^{z}a_f}\left(1.5^{a_e}-1\right)}{2^{\sum_{g=e}^{z}b_g}}=1$$

must have a unique solution for any $n_1$ such that $\left\{z,a_{1...z},b_{1...z}\in\mathbb{N^+}\right \}$

I am trying to find a connection between $(a_s,b_s)$ and $(a_{s+1},b_{s+1})$, such that an expression for the sum of these values for each iteration may be found, which would make the above expression more useful. Corrections and advice on whether this may or may not be possible would really be appreciated!

Thanks